0
$\begingroup$

Let $\Omega\subseteq \mathbb{R}^N$ be an open, bounded and connected set (it can be assumed with smooth boundary if necessary).

Consider $\phi:\Omega\to\mathbb{R}$, $\phi\in C^1(\overline{\Omega})$ (the Banach space of continuous functions on the closure of $\Omega$ having continuous partial derivatives on the closure of $\Omega$) satisfying:

$\bullet\ \Vert \phi\Vert_{C^1(\overline{\Omega})}\leq 1$

$\bullet\ \phi^{-1}(0)=\{x\in \Omega\ |\ \phi(x)=0\}\neq \emptyset$

$\bullet\ \nabla\phi(x)\neq 0$ for all $x\in\phi^{-1}(0)$.

Is it true that for each $\varepsilon>0$ there is a constant $\delta=\delta(\varepsilon)>0$ (depending just on $\varepsilon$) such that:

$$\lambda (\phi^{-1}(-\delta,\delta))<\varepsilon $$

?

I denote with $\lambda$ the Lebesgue measure on $\mathbb{R}^N$.

P.S. Our assumption guarantee that any level set $\phi=c$ has a null Lebesgue measure.

$\endgroup$
2
  • $\begingroup$ What exactly is $C^1(\bar{\Omega})$? In general the idea is that locally you may change the coordinates to $(x_1,\ldots,x_{i-1},\phi(x),x_{i+1},\ldots,x_n)$ by the implicit function theorem. $\endgroup$ – Fedor Petrov Feb 27 at 12:37
  • $\begingroup$ I mean that $\phi$ has contiunous partial derivatives on the closure of $\Omega$. $\endgroup$ – Bogdan Feb 27 at 12:40
2
$\begingroup$

The answer is no even for $N = 1$ and $\Omega = (-1,1)$, thus $\bar \Omega = [-1,1]$. Let $\phi_1 \colon [-1,1] \to [-1,1]$ be any increasing function (which may even be in $C^\infty$) with $\phi_1(0) = 0$ and $\phi_1'(x) > 0$ for all $x \in \bar \Omega$. Let $\phi_t := t \cdot \phi_1$. Then given $\epsilon > 0$ and $C > 0$ any fixed constant we get $\lambda(\phi_t^{-1}((-\epsilon,\epsilon)) = 2$ for $t < \epsilon$. Note that the assumptions are fulfilled for any $t \in (0,1)$.

$\endgroup$
4
  • $\begingroup$ Indeed. I forgot to mention that $\phi$ must be see as a function on $\Omega$ that have an extension to $\overline{\Omega}$. Can you provide an example with $\phi^{-1}(0)\cap \Omega\neq \emptyset$? $\endgroup$ – Bogdan Feb 27 at 13:23
  • $\begingroup$ Of course, simply let $\Omega = (-1,1)$ and $\phi_1 \colon [-1,1] \to [-1,1]$ any $C^\infty$-function with $\phi_1(0) = 0$ and $\phi_1'(x) > 0$ for all $x \in \bar \Omega$. I've edited the answer. $\endgroup$ – Dieter Kadelka Feb 27 at 13:29
  • $\begingroup$ Thanks a lot! I see now that my assumption is not correct. $\endgroup$ – Bogdan Feb 27 at 13:48
  • $\begingroup$ I have changed a little the conclusion, and now I think it's ok. $\endgroup$ – Bogdan Feb 28 at 7:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.